LeetCode 3536: Maximum Product of Two Digits ·

Table of Contents

Meta Data
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Difficulty: easy First Attempt: 2025-05-04

Submissions: Runtime Error ×4 → Wrong Answer ×2 → Accepted

Approach
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The answer is the product of the two largest digits (a digit appearing twice may be used twice). Sorting the digits and taking the last two is simplest; in one scan, keep max1 >= max2:

A digit > max1: the old max1 becomes max2 and the new digit becomes max1.
Otherwise, if > max2: update max2.

My Attempts
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Version	Problem
Four REs	`map(n, str)` had its arguments reversed; then type errors from comparing strings with the integer `0` and multiplying strings
Two WAs	Types fixed, but on a new maximum the old maximum was not moved to second place. For `n = 789` the answer is `9 × 8 = 72`; both returned 0. The first used `>=`, so the second 7 only overwrote first place, and `n = 707` returned 0 instead of 49
AC	Changed to `max1, max2 = s, max1`

Solution
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class Solution:
    def maxProduct(self, n: int) -> int:
        digits = sorted(int(c) for c in str(n))
        return digits[-1] * digits[-2]

Complexity
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Time O(d log d) for d digits (at most 10), space O(d).

Mistakes to Avoid
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Comparing strings with integers.
When tracking a top two, a new first place pushes the old first into second.

Interview Explanation
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The answer multiplies the two largest digits. I sort the digits and take the last two; in a single scan I would keep first and second place, demoting the old first whenever a new first appears. Time O(d).